3.200 \(\int \frac {\sin (a+\frac {b}{\sqrt {c+d x}})}{e+f x} \, dx\)

Optimal. Leaf size=276 \[ \frac {\sin \left (a-\frac {b \sqrt {f}}{\sqrt {c f-d e}}\right ) \text {Ci}\left (\frac {\sqrt {f} b}{\sqrt {c f-d e}}+\frac {b}{\sqrt {c+d x}}\right )}{f}+\frac {\sin \left (a+\frac {b \sqrt {f}}{\sqrt {c f-d e}}\right ) \text {Ci}\left (\frac {b \sqrt {f}}{\sqrt {c f-d e}}-\frac {b}{\sqrt {c+d x}}\right )}{f}-\frac {2 \sin (a) \text {Ci}\left (\frac {b}{\sqrt {c+d x}}\right )}{f}-\frac {\cos \left (a+\frac {b \sqrt {f}}{\sqrt {c f-d e}}\right ) \text {Si}\left (\frac {b \sqrt {f}}{\sqrt {c f-d e}}-\frac {b}{\sqrt {c+d x}}\right )}{f}+\frac {\cos \left (a-\frac {b \sqrt {f}}{\sqrt {c f-d e}}\right ) \text {Si}\left (\frac {\sqrt {f} b}{\sqrt {c f-d e}}+\frac {b}{\sqrt {c+d x}}\right )}{f}-\frac {2 \cos (a) \text {Si}\left (\frac {b}{\sqrt {c+d x}}\right )}{f} \]

[Out]

-cos(a+b*f^(1/2)/(c*f-d*e)^(1/2))*Si(b*f^(1/2)/(c*f-d*e)^(1/2)-b/(d*x+c)^(1/2))/f+cos(a-b*f^(1/2)/(c*f-d*e)^(1
/2))*Si(b*f^(1/2)/(c*f-d*e)^(1/2)+b/(d*x+c)^(1/2))/f-2*cos(a)*Si(b/(d*x+c)^(1/2))/f-2*Ci(b/(d*x+c)^(1/2))*sin(
a)/f+Ci(b*f^(1/2)/(c*f-d*e)^(1/2)+b/(d*x+c)^(1/2))*sin(a-b*f^(1/2)/(c*f-d*e)^(1/2))/f+Ci(b*f^(1/2)/(c*f-d*e)^(
1/2)-b/(d*x+c)^(1/2))*sin(a+b*f^(1/2)/(c*f-d*e)^(1/2))/f

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Rubi [A]  time = 1.20, antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {3431, 3303, 3299, 3302, 3345} \[ \frac {\sin \left (a-\frac {b \sqrt {f}}{\sqrt {c f-d e}}\right ) \text {CosIntegral}\left (\frac {b \sqrt {f}}{\sqrt {c f-d e}}+\frac {b}{\sqrt {c+d x}}\right )}{f}+\frac {\sin \left (a+\frac {b \sqrt {f}}{\sqrt {c f-d e}}\right ) \text {CosIntegral}\left (\frac {b \sqrt {f}}{\sqrt {c f-d e}}-\frac {b}{\sqrt {c+d x}}\right )}{f}-\frac {2 \sin (a) \text {CosIntegral}\left (\frac {b}{\sqrt {c+d x}}\right )}{f}-\frac {\cos \left (a+\frac {b \sqrt {f}}{\sqrt {c f-d e}}\right ) \text {Si}\left (\frac {b \sqrt {f}}{\sqrt {c f-d e}}-\frac {b}{\sqrt {c+d x}}\right )}{f}+\frac {\cos \left (a-\frac {b \sqrt {f}}{\sqrt {c f-d e}}\right ) \text {Si}\left (\frac {\sqrt {f} b}{\sqrt {c f-d e}}+\frac {b}{\sqrt {c+d x}}\right )}{f}-\frac {2 \cos (a) \text {Si}\left (\frac {b}{\sqrt {c+d x}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/Sqrt[c + d*x]]/(e + f*x),x]

[Out]

(-2*CosIntegral[b/Sqrt[c + d*x]]*Sin[a])/f + (CosIntegral[(b*Sqrt[f])/Sqrt[-(d*e) + c*f] + b/Sqrt[c + d*x]]*Si
n[a - (b*Sqrt[f])/Sqrt[-(d*e) + c*f]])/f + (CosIntegral[(b*Sqrt[f])/Sqrt[-(d*e) + c*f] - b/Sqrt[c + d*x]]*Sin[
a + (b*Sqrt[f])/Sqrt[-(d*e) + c*f]])/f - (2*Cos[a]*SinIntegral[b/Sqrt[c + d*x]])/f - (Cos[a + (b*Sqrt[f])/Sqrt
[-(d*e) + c*f]]*SinIntegral[(b*Sqrt[f])/Sqrt[-(d*e) + c*f] - b/Sqrt[c + d*x]])/f + (Cos[a - (b*Sqrt[f])/Sqrt[-
(d*e) + c*f]]*SinIntegral[(b*Sqrt[f])/Sqrt[-(d*e) + c*f] + b/Sqrt[c + d*x]])/f

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3345

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*Sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[ExpandIntegrand[Sin[c +
d*x], x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && ILtQ[p, 0] && IGtQ[n, 0] && (EqQ[n, 2] || EqQ
[p, -1]) && IntegerQ[m]

Rule 3431

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Dist[1/(n*f), Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x])^p, x^(1/n - 1)*(g - (e*h)/f + (h*x^(1/n))/f)^m,
x], x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && IGtQ[p, 0] && IntegerQ[1/n]

Rubi steps

\begin {align*} \int \frac {\sin \left (a+\frac {b}{\sqrt {c+d x}}\right )}{e+f x} \, dx &=-\frac {2 \operatorname {Subst}\left (\int \left (\frac {d \sin (a+b x)}{f x}+\frac {d (-d e+c f) x \sin (a+b x)}{f \left (f+(d e-c f) x^2\right )}\right ) \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{d}\\ &=-\frac {2 \operatorname {Subst}\left (\int \frac {\sin (a+b x)}{x} \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{f}+\frac {(2 (d e-c f)) \operatorname {Subst}\left (\int \frac {x \sin (a+b x)}{f+(d e-c f) x^2} \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{f}\\ &=\frac {(2 (d e-c f)) \operatorname {Subst}\left (\int \left (-\frac {\sqrt {-d e+c f} \sin (a+b x)}{2 (d e-c f) \left (\sqrt {f}-\sqrt {-d e+c f} x\right )}+\frac {\sqrt {-d e+c f} \sin (a+b x)}{2 (d e-c f) \left (\sqrt {f}+\sqrt {-d e+c f} x\right )}\right ) \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{f}-\frac {(2 \cos (a)) \operatorname {Subst}\left (\int \frac {\sin (b x)}{x} \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{f}-\frac {(2 \sin (a)) \operatorname {Subst}\left (\int \frac {\cos (b x)}{x} \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{f}\\ &=-\frac {2 \text {Ci}\left (\frac {b}{\sqrt {c+d x}}\right ) \sin (a)}{f}-\frac {2 \cos (a) \text {Si}\left (\frac {b}{\sqrt {c+d x}}\right )}{f}-\frac {\sqrt {-d e+c f} \operatorname {Subst}\left (\int \frac {\sin (a+b x)}{\sqrt {f}-\sqrt {-d e+c f} x} \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{f}+\frac {\sqrt {-d e+c f} \operatorname {Subst}\left (\int \frac {\sin (a+b x)}{\sqrt {f}+\sqrt {-d e+c f} x} \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{f}\\ &=-\frac {2 \text {Ci}\left (\frac {b}{\sqrt {c+d x}}\right ) \sin (a)}{f}-\frac {2 \cos (a) \text {Si}\left (\frac {b}{\sqrt {c+d x}}\right )}{f}+\frac {\left (\sqrt {-d e+c f} \cos \left (a-\frac {b \sqrt {f}}{\sqrt {-d e+c f}}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {b \sqrt {f}}{\sqrt {-d e+c f}}+b x\right )}{\sqrt {f}+\sqrt {-d e+c f} x} \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{f}+\frac {\left (\sqrt {-d e+c f} \cos \left (a+\frac {b \sqrt {f}}{\sqrt {-d e+c f}}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {b \sqrt {f}}{\sqrt {-d e+c f}}-b x\right )}{\sqrt {f}-\sqrt {-d e+c f} x} \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{f}+\frac {\left (\sqrt {-d e+c f} \sin \left (a-\frac {b \sqrt {f}}{\sqrt {-d e+c f}}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {b \sqrt {f}}{\sqrt {-d e+c f}}+b x\right )}{\sqrt {f}+\sqrt {-d e+c f} x} \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{f}-\frac {\left (\sqrt {-d e+c f} \sin \left (a+\frac {b \sqrt {f}}{\sqrt {-d e+c f}}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {b \sqrt {f}}{\sqrt {-d e+c f}}-b x\right )}{\sqrt {f}-\sqrt {-d e+c f} x} \, dx,x,\frac {1}{\sqrt {c+d x}}\right )}{f}\\ &=-\frac {2 \text {Ci}\left (\frac {b}{\sqrt {c+d x}}\right ) \sin (a)}{f}+\frac {\text {Ci}\left (\frac {b \sqrt {f}}{\sqrt {-d e+c f}}+\frac {b}{\sqrt {c+d x}}\right ) \sin \left (a-\frac {b \sqrt {f}}{\sqrt {-d e+c f}}\right )}{f}+\frac {\text {Ci}\left (\frac {b \sqrt {f}}{\sqrt {-d e+c f}}-\frac {b}{\sqrt {c+d x}}\right ) \sin \left (a+\frac {b \sqrt {f}}{\sqrt {-d e+c f}}\right )}{f}-\frac {2 \cos (a) \text {Si}\left (\frac {b}{\sqrt {c+d x}}\right )}{f}-\frac {\cos \left (a+\frac {b \sqrt {f}}{\sqrt {-d e+c f}}\right ) \text {Si}\left (\frac {b \sqrt {f}}{\sqrt {-d e+c f}}-\frac {b}{\sqrt {c+d x}}\right )}{f}+\frac {\cos \left (a-\frac {b \sqrt {f}}{\sqrt {-d e+c f}}\right ) \text {Si}\left (\frac {b \sqrt {f}}{\sqrt {-d e+c f}}+\frac {b}{\sqrt {c+d x}}\right )}{f}\\ \end {align*}

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Mathematica [F]  time = 15.69, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt {c+d x}}\right )}{e+f x} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sin[a + b/Sqrt[c + d*x]]/(e + f*x),x]

[Out]

Integrate[Sin[a + b/Sqrt[c + d*x]]/(e + f*x), x]

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fricas [C]  time = 0.80, size = 320, normalized size = 1.16 \[ \frac {2 i \, {\rm Ei}\left (\frac {i \, b}{\sqrt {d x + c}}\right ) e^{\left (i \, a\right )} - 2 i \, {\rm Ei}\left (-\frac {i \, b}{\sqrt {d x + c}}\right ) e^{\left (-i \, a\right )} - i \, {\rm Ei}\left (-\frac {2 \, \sqrt {\frac {b^{2} f}{d e - c f}} {\left (d x + c\right )} - 2 i \, \sqrt {d x + c} b}{2 \, {\left (d x + c\right )}}\right ) e^{\left (i \, a + \sqrt {\frac {b^{2} f}{d e - c f}}\right )} - i \, {\rm Ei}\left (\frac {2 \, \sqrt {\frac {b^{2} f}{d e - c f}} {\left (d x + c\right )} + 2 i \, \sqrt {d x + c} b}{2 \, {\left (d x + c\right )}}\right ) e^{\left (i \, a - \sqrt {\frac {b^{2} f}{d e - c f}}\right )} + i \, {\rm Ei}\left (-\frac {2 \, \sqrt {\frac {b^{2} f}{d e - c f}} {\left (d x + c\right )} + 2 i \, \sqrt {d x + c} b}{2 \, {\left (d x + c\right )}}\right ) e^{\left (-i \, a + \sqrt {\frac {b^{2} f}{d e - c f}}\right )} + i \, {\rm Ei}\left (\frac {2 \, \sqrt {\frac {b^{2} f}{d e - c f}} {\left (d x + c\right )} - 2 i \, \sqrt {d x + c} b}{2 \, {\left (d x + c\right )}}\right ) e^{\left (-i \, a - \sqrt {\frac {b^{2} f}{d e - c f}}\right )}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)^(1/2))/(f*x+e),x, algorithm="fricas")

[Out]

1/2*(2*I*Ei(I*b/sqrt(d*x + c))*e^(I*a) - 2*I*Ei(-I*b/sqrt(d*x + c))*e^(-I*a) - I*Ei(-1/2*(2*sqrt(b^2*f/(d*e -
c*f))*(d*x + c) - 2*I*sqrt(d*x + c)*b)/(d*x + c))*e^(I*a + sqrt(b^2*f/(d*e - c*f))) - I*Ei(1/2*(2*sqrt(b^2*f/(
d*e - c*f))*(d*x + c) + 2*I*sqrt(d*x + c)*b)/(d*x + c))*e^(I*a - sqrt(b^2*f/(d*e - c*f))) + I*Ei(-1/2*(2*sqrt(
b^2*f/(d*e - c*f))*(d*x + c) + 2*I*sqrt(d*x + c)*b)/(d*x + c))*e^(-I*a + sqrt(b^2*f/(d*e - c*f))) + I*Ei(1/2*(
2*sqrt(b^2*f/(d*e - c*f))*(d*x + c) - 2*I*sqrt(d*x + c)*b)/(d*x + c))*e^(-I*a - sqrt(b^2*f/(d*e - c*f))))/f

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (a + \frac {b}{\sqrt {d x + c}}\right )}{f x + e}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)^(1/2))/(f*x+e),x, algorithm="giac")

[Out]

integrate(sin(a + b/sqrt(d*x + c))/(f*x + e), x)

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maple [A]  time = 0.06, size = 438, normalized size = 1.59 \[ -2 b^{2} \left (\frac {\Si \left (\frac {b}{\sqrt {d x +c}}\right ) \cos \relax (a )+\Ci \left (\frac {b}{\sqrt {d x +c}}\right ) \sin \relax (a )}{b^{2} f}-\frac {\Si \left (\frac {b}{\sqrt {d x +c}}+a -\frac {a c f -a d e +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{c f -d e}\right ) \cos \left (\frac {a c f -a d e +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{c f -d e}\right )+\Ci \left (\frac {b}{\sqrt {d x +c}}+a -\frac {a c f -a d e +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{c f -d e}\right ) \sin \left (\frac {a c f -a d e +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{c f -d e}\right )}{2 b^{2} f}-\frac {\Si \left (\frac {b}{\sqrt {d x +c}}+a +\frac {-a c f +a d e +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{c f -d e}\right ) \cos \left (\frac {-a c f +a d e +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{c f -d e}\right )-\Ci \left (\frac {b}{\sqrt {d x +c}}+a +\frac {-a c f +a d e +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{c f -d e}\right ) \sin \left (\frac {-a c f +a d e +\sqrt {b^{2} c \,f^{2}-b^{2} d e f}}{c f -d e}\right )}{2 b^{2} f}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b/(d*x+c)^(1/2))/(f*x+e),x)

[Out]

-2*b^2*(1/b^2/f*(Si(b/(d*x+c)^(1/2))*cos(a)+Ci(b/(d*x+c)^(1/2))*sin(a))-1/2/b^2/f*(Si(b/(d*x+c)^(1/2)+a-(a*c*f
-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*cos((a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))+Ci(b/(
d*x+c)^(1/2)+a-(a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*sin((a*c*f-a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/
2))/(c*f-d*e)))-1/2/b^2/f*(Si(b/(d*x+c)^(1/2)+a+(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))*cos((-a*
c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))-Ci(b/(d*x+c)^(1/2)+a+(-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/
2))/(c*f-d*e))*sin((-a*c*f+a*d*e+(b^2*c*f^2-b^2*d*e*f)^(1/2))/(c*f-d*e))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (a + \frac {b}{\sqrt {d x + c}}\right )}{f x + e}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)^(1/2))/(f*x+e),x, algorithm="maxima")

[Out]

integrate(sin(a + b/sqrt(d*x + c))/(f*x + e), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sin \left (a+\frac {b}{\sqrt {c+d\,x}}\right )}{e+f\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/(c + d*x)^(1/2))/(e + f*x),x)

[Out]

int(sin(a + b/(c + d*x)^(1/2))/(e + f*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin {\left (a + \frac {b}{\sqrt {c + d x}} \right )}}{e + f x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)**(1/2))/(f*x+e),x)

[Out]

Integral(sin(a + b/sqrt(c + d*x))/(e + f*x), x)

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